Seed is 4m/s 3 sec later at the bottom of the slope it's speed is 22m/s what is the average acceleration
Answers
U2 = 22 m/s
t1 = 0
t2 = 3 sec.
α=
Related Questions
Endothermic reactions give off heat as they proceed. a. True b. False
Powerful eruptions on the sun’s surface are called ________. a. sunspots b. solar flares c. coronas d. supernovae
Using the terms humidity and dew point, explain why you might find small droplets of water on your lawn in the morning.
What is the charge of the nucleus
Answers
Answer: a
Explanation: Hope it helps
Answer:
A
Explanation:
I think its a because A would be heavier making it go down faster than B. its like when you drop a father compared to a ball, the feather is lighter so it will go down slower than the ball because it has more mass
Answers
Answer: E3 > E1 > E2 = E4
Explanation:
Electric current passes through the _____ and into the electromagnets in an electric motor.
A motor turns _____ energy into _____ energy.
Answers
Answer:
A _commutator_ is used in a motor to switch the direction of the magnetic field created by the current.
The rotating part of a motor that holds the electromagnets is called the __armature___.
Electric current passes through the _brushes_ and into the electromagnets in an electric motor.
A motor turns _electrical_ energy into _mechanical_ energy.
Explanation:
A commutator, which is a split ring rotary switching device, reverses the direction of the current between the external circuit and the rotor. Reversing the current reverses the magnetic field.
The armature comprises the rotating part of the motor and the electromagnets
A brush is the electrical contact for conducting current through the moving and stationary parts of an electric motor
An electric motor turns electrical energy into mechanical energy.
Answers
Answer:
A) α = -1.228 rev/min²
B) 7980 revolutions
C) α_t = -8.57 x 10^(-4) m/s²
D) α = 21.5 m/s²
Explanation:
A) Using first equation of motion, we have;
ω = ω_o + αt
Where,
ω_o is initial angular velocity
α is angular acceleration
t is time the flywheel take to slow down to rest.
We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min
Thus,
0 = 140 + 114α
α = -140/114
α = -1.228 rev/min²
B) the number of revolutions would be given by the equation of motion;
S = (ω_o)t + (1/2)αt²
S = 140(114) - (1/2)(1.228)(114)²
S ≈ 7980 revolutions
C) we want to find tangential component of the velocity with r = 40cm = 0.4m
We will need to convert the angular acceleration to rad/s²
Thus,
α = -1.228 x (2π/60²) = - 0.0021433 rad/s²
Now, formula for tangential acceleration is;
α_t = α x r
α_t = - 0.0021433 x 0.4
α_t = -8.57 x 10^(-4) m/s²
D) we are told that the angular velocity is now 70 rev/min.
Let's convert it to rad/s;
ω = 70 x (2π/60) = 7.33 rad/s
So, radial angular acceleration is;
α_r = ω²r = 7.33² x 0.4
α_r = 21.49 m/s²
Thus, magnitude of total linear acceleration is;
α = √((α_t)² + (α_r)²)
α = √((-8.57 x 10^(-4))² + (21.49)²)
α = √461.82
α = 21.5 m/s²
At which landing site would the lander have the greatest amount of gravitational potential energy?
A. W
B. X
C. Y
D. Z
Answers
Answer:
B. X.
Explanation:
To determine which landing site would have the greatest amount of gravitational potential energy, we need to consider the height above the surface and the acceleration due to gravity at each site.
Gravitational potential energy is given by the formula:
Gravitational potential energy = mass x acceleration due to gravity x height
In this case, the mass of the lander is not provided, but since it is the same for all the sites, we can ignore it for the purpose of comparison. Therefore, we only need to consider the acceleration due to gravity and the height above the surface.
Looking at the table, we can see that at site X, the height above the surface is 16 meters, and the acceleration due to gravity is 3.7 meters per second squared. This means that at site X, the lander would have the highest amount of gravitational potential energy compared to the other sites.
i hoped this helped ! ⚫w⚫
b. False